A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is `k`. If radius of the ball be `R`, then the fraction of total energy associated with its rotation will be.

Options

A. `(K^(2))/(K^(2)+R^(2))`
B. `(R^(2))/(K^(2)+R^(2))`
C. `(K^(2)+R^(2))/(R^(2))`
D. `(K^(2))/(R^(2))`

Answer

A

Explanation

`v=Romega` in case of pure rolling.
`therefore" "K_("Rotational")=(1)/(2)l omega^(2)=(1)/(2)mK^(2)+(v^(2))/(R^(2))`
`"where, "l=mK^(2)`
`"K= Radius of gyration"`
`K_("Total")=K_("Rot")+K_("Trans")`
`=(1)/(2)mK^(2)(v^(2))/(R^(2))+(1)/(2)mv^(2)`
Thus, `(K_(R))/(K_("Total"))=(K^(2))/(K^(2)+R^(2))`